Math for Assembly
Submit this sheet with all calculations and all supporting work.
Convert the following decimal numbers to binary:
(a) 125 2 125 ÷ 2 r e m a i n d e r 1 62 ÷ 2 r e m a i n d e r 0 31 ÷ 2 r e m a i n d e r 1 15 ÷ 2 r e m a i n d e r 1 7 ÷ 2 r e m a i n d e r 1 3 ÷ 2 r e m a i n d e r 1 1 ÷ 2 r e m a i n d e r 1 N o t e : 2 6 + 2 5 + 2 4 + 2 3 + 2 2 + 2 0 64 + 32 + 16 + 8 + 4 + 1 = 125 ∴ 125 in binary form is 111 1101 (b) 42 2 42 ÷ 2 r e m a i n d e r 0 21 ÷ 2 r e m a i n d e r 1 10 ÷ 2 r e m a i n d e r 0 5 ÷ 2 r e m a i n d e r 1 2 ÷ 2 r e m a i n d e r 0 1 ÷ 2 r e m a i n d e r 1 N o t e : 2 5 + 2 3 + 2 1 32 + 8 + 2 = 125 ∴ 42 in binary form is 10 1010
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 125}}\qquad &\\
2&\sqrt{125} & \div2\ remainder\ 1\\
&62 & \div2\ remainder\ 0\\
&31 & \div2\ remainder\ 1\\
&15 & \div2\ remainder\ 1\\
&7 & \div2\ remainder\ 1\\
&3 & \div2\ remainder\ 1\\
&1 & \div2\ remainder\ 1\\
Note: &\quad 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^0 \\
&\quad 64 + 32 + 16 + 8 + 4 + 1 = 125\\
\therefore &\quad\text{\textcolor{khaki}{125 in binary form is 111 1101}}\\[2em]
\textrm{\textcolor{skyblue}{(b) 42}}\qquad &\\
2&\sqrt{42} & \div2\ remainder\ 0\\
&21 & \div2\ remainder\ 1\\
&10 & \div2\ remainder\ 0\\
&5 & \div2\ remainder\ 1\\
&2 & \div2\ remainder\ 0\\
&1 & \div2\ remainder\ 1\\
Note: &\quad 2^5 + 2^3 + 2^1 \\
&\quad 32 + 8 + 2 = 125\\
\therefore &\quad\text{\textcolor{khaki}{42 in binary form is 10 1010}}
\end{aligned}
(a) 125 2 N o t e : ∴ (b) 42 2 N o t e : ∴ 125 62 31 15 7 3 1 2 6 + 2 5 + 2 4 + 2 3 + 2 2 + 2 0 64 + 32 + 16 + 8 + 4 + 1 = 125 125 in binary form is 111 1101 42 21 10 5 2 1 2 5 + 2 3 + 2 1 32 + 8 + 2 = 125 42 in binary form is 10 1010 ÷ 2 re main d er 1 ÷ 2 re main d er 0 ÷ 2 re main d er 1 ÷ 2 re main d er 1 ÷ 2 re main d er 1 ÷ 2 re main d er 1 ÷ 2 re main d er 1 ÷ 2 re main d er 0 ÷ 2 re main d er 1 ÷ 2 re main d er 0 ÷ 2 re main d er 1 ÷ 2 re main d er 0 ÷ 2 re main d er 1
Convert the following binary numbers to decimal:
2 15 2 14 2 13 2 12 2 11 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 32768 16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 1 0 0 1 0 0 1 0 1 1 0 1 1 0 0 0 2 15 2 14 2 13 2 12 2 11 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 32,768 + 4,096 + 512 + 128 + 64 + 16 + 8 = 37, 592 ∴ 1001 0010 1101 1000 in decimal form is 37, 592 1 0 1 1 1 0 1 0 1 0 1 0 1 1 1 1 2 15 2 14 2 13 2 12 2 11 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 32768 16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 32,768 + 8,192 + 4096 + 2048 + 512 + 128 + 32 + 8 + 4 + 2 + 1 = 47, 791 ∴ 1011 1010 1010 1111 in decimal form is 47, 791
\begin{matrix*}[c]
2^{15} & 2^{14} & 2^{13} & 2^{12} & 2^{11} & 2^{10} & 2^9 & 2^8 & 2^7 & 2^6 & 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^0 &\\
32768 & 16384 & 8192 & 4096 & 2048 & 1024 & 512 & 256 & 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1 &\\\\
1&0&0&1& 0&0&1&0& 1&1&0&1& 1&0&0&0&\\
2^{15} & \phantom{ 2^{14} } & \phantom{ 2^{13} } & 2^{12} & \phantom{ 2^{11} } & \phantom{ 2^{10} } & 2^9 & \phantom{ 2^8 } & 2^7 & 2^6 & \phantom{ 2^5 } & 2^4 & 2^3 & \phantom{ 2^2 } & \phantom{ 2^1 } & \phantom{ 2^0 } &\\
\end{matrix*}\\[2em]
\textsf{32,768 + 4,096 + 512 + 128 + 64 + 16 + 8 = 37, 592} \\
\therefore \quad\text{\textcolor{khaki}{1001 0010 1101 1000 in decimal form is 37, 592}}\\[2em]
\begin{matrix*}[c]
1&0&1&1 &1&0&1&0 &1&0&1&0 &1&1&1&1&\\
2^{15} & \phantom{ 2^{14} } & 2^{13} & 2^{12} & 2^{11} & \phantom{ 2^{10} } & 2^9 & \phantom{ 2^8 } & 2^7 & \phantom{ 2^6 } & 2^5 & \phantom{ 2^4 } & 2^3 & 2^2 & 2^1 & 2^0 &\\
32768 & \phantom{ 16384 } & 8192 & 4096 & 2048 & \phantom{ 1024 } & 512 & \phantom{ 256 } & 128 & \phantom{ 64 } & 32 & \phantom{ 16 } & 8 & 4 & 2 & 1 &\\
\end{matrix*}\\[2em]
\textsf{32,768 + 8,192 + 4096 + 2048 + 512 + 128 + 32 + 8 + 4 + 2 + 1 = 47, 791} \\
\therefore \quad\text{\textcolor{khaki}{1011 1010 1010 1111 in decimal form is 47, 791}}\\[2em]
2 15 32768 1 2 15 2 14 16384 0 2 14 2 13 8192 0 2 13 2 12 4096 1 2 12 2 11 2048 0 2 11 2 10 1024 0 2 10 2 9 512 1 2 9 2 8 256 0 2 8 2 7 128 1 2 7 2 6 64 1 2 6 2 5 32 0 2 5 2 4 16 1 2 4 2 3 8 1 2 3 2 2 4 0 2 2 2 1 2 0 2 1 2 0 1 0 2 0 32,768 + 4,096 + 512 + 128 + 64 + 16 + 8 = 37, 592 ∴ 1001 0010 1101 1000 in decimal form is 37, 592 1 2 15 32768 0 2 14 16384 1 2 13 8192 1 2 12 4096 1 2 11 2048 0 2 10 1024 1 2 9 512 0 2 8 256 1 2 7 128 0 2 6 64 1 2 5 32 0 2 4 16 1 2 3 8 1 2 2 4 1 2 1 2 1 2 0 1 32,768 + 8,192 + 4096 + 2048 + 512 + 128 + 32 + 8 + 4 + 2 + 1 = 47, 791 ∴ 1011 1010 1010 1111 in decimal form is 47, 791
Convert the following decimal numbers to hexadecimal:
(a) 925 16 925 ÷ 16 r e m a i n d e r 13 o r D 57 ÷ 16 r e m a i n d e r 9 3 ÷ 16 r e m a i n d e r 3 N o t e : 1 6 2 × 3 + 1 6 2 × 9 + 1 6 0 × 3 768 + 144 + 3 = 125 ∴ 925 in hexadecimal form is 39D (b) 104 16 104 ÷ 16 r e m a i n d e r 8 6 ÷ 16 r e m a i n d e r 6 N o t e : 1 6 1 × 6 + 1 6 0 × 8 96 + 8 = 104 ∴ 104 in hexadecimal form is 68
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 925}}\qquad &\\
16&\sqrt{925} & \div 16\ remainder\ 13\ or\ D\\
&57 & \div 16\ remainder\ 9\\
&3 & \div 16\ remainder\ 3\\
Note: &\quad 16^2\times 3 + 16^2 \times 9 + 16^0 \times 3 \\
&\quad 768 + 144 + 3 = 125\\
\therefore &\quad\text{\textcolor{khaki}{925 in hexadecimal form is 39D}}\\[2em]
\textrm{\textcolor{skyblue}{(b) 104}}\qquad &\\
16&\sqrt{104} & \div16\ remainder\ 8\\
&6 & \div16\ remainder\ 6\\
Note: &\quad 16^1 \times 6 + 16^0 \times 8 \\
&\quad 96 + 8 = 104\\
\therefore &\quad\text{\textcolor{khaki}{104 in hexadecimal form is 68}}
\end{aligned}
(a) 925 16 N o t e : ∴ (b) 104 16 N o t e : ∴ 925 57 3 1 6 2 × 3 + 1 6 2 × 9 + 1 6 0 × 3 768 + 144 + 3 = 125 925 in hexadecimal form is 39D 104 6 1 6 1 × 6 + 1 6 0 × 8 96 + 8 = 104 104 in hexadecimal form is 68 ÷ 16 re main d er 13 or D ÷ 16 re main d er 9 ÷ 16 re main d er 3 ÷ 16 re main d er 8 ÷ 16 re main d er 6
Convert the following hexadecimal numbers to decimal:
(a) 0 x 5F53DA ( 5 F 53 D A ) 16 = 5 × 1 6 5 = 5 , 242 , 880 + 15 × 1 6 4 = 983 , 040 + 5 × 1 6 3 = 20 , 480 + 3 × 1 6 2 = 768 + 13 × 1 6 1 = 208 + 10 × 1 6 0 = + 10 ∴ 0 x 5F53DA in decimal form is 6 , 247 , 386 ‾ (a) 0 x 54ABC2 ( 54 A B C 2 ) 16 = 5 × 1 6 5 = 5 , 242 , 880 + 4 × 1 6 4 = 262 , 144 + 10 × 1 6 3 = 40 , 960 + 11 × 1 6 2 = 2 , 816 + 12 × 1 6 1 = 192 + 2 × 1 6 0 = + 2 ∴ 0 x 54ABC2 in decimal form is 5 , 548 , 994 ‾
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 0 x 5F53DA}}\qquad\qquad&\\
(5F53DA)_{16}=\qquad &5&\times 16^5& =& 5,242,880&\\
+\qquad &15&\times 16^4& =& 983,040&\\
+\qquad &5&\times 16^3& =& 20,480&\\
+\qquad &3&\times 16^2& =& 768&\\
+\qquad &13&\times 16^1& =& 208&\\
+\qquad &10&\times 16^0& =&+\qquad 10&\\
&&&\therefore& \quad\text{\textcolor{khaki}{0 x 5F53DA in decimal form is\qquad $\overline{6,247,386}$}}& \\[2em]
\textrm{\textcolor{skyblue}{(a) 0 x 54ABC2}}\qquad\qquad&\\
(54ABC2)_{16}=\qquad &5&\times 16^5& =& 5,242,880&\\
+\qquad &4&\times 16^4& =& 262,144&\\
+\qquad &10&\times 16^3& =& 40,960&\\
+\qquad &11&\times 16^2& =& 2,816&\\
+\qquad &12&\times 16^1& =& 192&\\
+\qquad &2&\times 16^0& =&+\qquad 2&\\
&&&\therefore& \quad\text{\textcolor{khaki}{0 x 54ABC2 in decimal form is\qquad $\overline{5,548,994}$}}& \\[2em]
\end{aligned}
(a) 0 x 5F53DA ( 5 F 53 D A ) 16 = + + + + + (a) 0 x 54ABC2 ( 54 A BC 2 ) 16 = + + + + + 5 15 5 3 13 10 5 4 10 11 12 2 × 1 6 5 × 1 6 4 × 1 6 3 × 1 6 2 × 1 6 1 × 1 6 0 × 1 6 5 × 1 6 4 × 1 6 3 × 1 6 2 × 1 6 1 × 1 6 0 = = = = = = ∴ = = = = = = ∴ 5 , 242 , 880 983 , 040 20 , 480 768 208 + 10 0 x 5F53DA in decimal form is 6 , 247 , 386 5 , 242 , 880 262 , 144 40 , 960 2 , 816 192 + 2 0 x 54ABC2 in decimal form is 5 , 548 , 994
Convert the following binary numbers to hexadecimal:
(a) 1011 1000 0001 0000 1011 ⏟ 11 ⏟ B 1000 ⏟ 8 0001 ⏟ 1 0000 ⏟ 0 ∴ 1011 1000 0001 0000 in hexadecimal is B810 (b) 1010 0100 0011 1100 1010 ⏟ 10 ⏟ A 0100 ⏟ 4 0011 ⏟ 3 1100 ⏟ 12 ⏟ C ∴ 1010 0100 0011 1100 in hexadecimal is A43C
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 1011 1000 0001 0000}}\qquad\qquad&\\[1em]
\underbrace{\underbrace{1011}_{\text{11}}}_{\text{B}}\
\underbrace{1000}_{\text{8}}\
\underbrace{0001}_{\text{1}}\
\underbrace{0000}_{\text{0}}\ &\\\\
\therefore\text{\textcolor{khaki}{1011 1000 0001 0000 in hexadecimal is B810}}\\[2em]
\textrm{\textcolor{skyblue}{(b) 1010 0100 0011 1100}}\qquad\qquad&\\[1em]
\underbrace{ \underbrace{1010}_{\text{10}} }_{\text{A}}\
\underbrace{0100}_{\text{4}}\
\underbrace{0011}_{\text{3}}\
\underbrace{ \underbrace{1100}_{\text{12}} }_{\text{C}}\ &\\\\
\therefore\text{\textcolor{khaki}{1010 0100 0011 1100 in hexadecimal is A43C}}
\end{aligned}
(a) 1011 1000 0001 0000 B 11 1011 8 1000 1 0001 0 0000 ∴ 1011 1000 0001 0000 in hexadecimal is B810 (b) 1010 0100 0011 1100 A 10 1010 4 0100 3 0011 C 12 1100 ∴ 1010 0100 0011 1100 in hexadecimal is A43C
Decimal
Hexadecimal
Binary
0
0
0000
1
1
0001
2
2
0010
3
3
0011
4
4
0100
5
5
0101
6
6
0110
7
7
0111
8
8
1000
9
9
1001
10
A
1010
11
B
1011
12
C
1100
13
D
1101
14
E
1110
15
F
1111
Convert the following hexadecimal numbers to binary:
(a) 0 × 5CD6 5 ⏟ 0101 C ⏟ 12 ⏟ 1100 D ⏟ 13 ⏟ 1101 6 ⏟ 0110 ∴ 0 × 5CD6 in hexadecimal to binary is 101 1100 1101 0110 (b) 0 × 2BCD 2 ⏟ 0010 B ⏟ 11 ⏟ 1011 C ⏟ 12 ⏟ 1100 D ⏟ 13 ⏟ 1101 ∴ 0 × 2BCD in hexadecimal to binary is 10 1011 1100 1101
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 0 $\times$ 5CD6}}\qquad\qquad&\\[1em]
\underbrace{5}_{\text{0101}}\
\underbrace{\underbrace{C}_{\text{12}}}_{\text{1100}}\
\underbrace{\underbrace{D}_{\text{13}}}_{\text{1101}}\
\underbrace{6}_{\text{0110}}\ &\\\\
\therefore\text{\textcolor{khaki}{0 $\times$ 5CD6 in hexadecimal to binary is 101 1100 1101 0110}}\\[2em]
\textrm{\textcolor{skyblue}{(b) 0 $\times$ 2BCD}}\qquad\qquad&\\[1em]
\underbrace{2}_{\text{0010}}\
\underbrace{ \underbrace{B}_{\text{11}} }_{\text{1011}}\
\underbrace{ \underbrace{C}_{\text{12}} }_{\text{1100}}\
\underbrace{ \underbrace{D}_{\text{13}} }_{\text{1101}}\\\\
\therefore\text{\textcolor{khaki}{0 $\times$ 2BCD in hexadecimal to binary is 10 1011 1100 1101}}\\[2em]
\end{aligned}
(a) 0 × 5CD6 0101 5 1100 12 C 1101 13 D 0110 6 ∴ 0 × 5CD6 in hexadecimal to binary is 101 1100 1101 0110 (b) 0 × 2BCD 0010 2 1011 11 B 1100 12 C 1101 13 D ∴ 0 × 2BCD in hexadecimal to binary is 10 1011 1100 1101
Calculate the 2’s complement of each of the following numbers:
(a) 1000 0101 ( 1’s complement ) 0111 1010 + 0000 0001 ( 2’s complement ) 0111 1011 ‾ (a) 1001 1101 ( 1’s complement ) 0110 0010 + 0000 0001 ( 2’s complement ) 0110 0011 ‾
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 1000 0101}}\qquad\qquad&\\[1em]
\text{( 1's complement ) }\qquad 0111\ 1010&\\
+0000\ 0001&\\
\text{\textcolor{khaki}{( 2's complement )\qquad $\overline{0111\ 1011}$}}& \\[2em]
\textrm{\textcolor{skyblue}{(a) 1001 1101}}\qquad\qquad&\\[1em]
\text{( 1's complement ) }\qquad 0110\ 0010&\\
+0000\ 0001&\\
\text{\textcolor{khaki}{( 2's complement )\qquad $\overline{0110\ 0011}$}}& \\[2em]
\end{aligned}
(a) 1000 0101 ( 1’s complement ) 0111 1010 + 0000 0001 ( 2’s complement ) 0111 1011 (a) 1001 1101 ( 1’s complement ) 0110 0010 + 0000 0001 ( 2’s complement ) 0110 0011
Perform the following signed operations in binary. Do not convert the numbers.
(a) 1101 1000 + 1000 1100 11 0 1 1 1 1000 + 1000 1100 1 0110 0100 ‾ (b) 1110 1010 – 0010 0100 1110 1 0 0 10 10 − 0010 0100 1100 0110 ‾
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 1101 1000 + 1000 1100}}\qquad\qquad&\\[1em]
11\overset{1}{0}\overset{1}{1} \ 1000\\
+\quad1000\ 1100\\
\text{\textcolor{khaki}{ $\overline{1\ 0110\ 0100}$}}& \\[2em]
\textrm{\textcolor{skyblue}{(b) 1110 1010 – 0010 0100}}\qquad\qquad&\\[1em]
1110\ \overset{0}{\cancel{1}}\overset{10}{0}10\\
-\quad0010\ 0100\\
\text{\textcolor{khaki}{ $\overline{1100\ 0110}$}}& \\[2em]
\end{aligned}
(a) 1101 1000 + 1000 1100 11 0 1 1 1 1000 + 1000 1100 1 0110 0100 (b) 1110 1010 – 0010 0100 1110 1 0 0 10 10 − 0010 0100 1100 0110
Perform the following signed operations in hexadecimal. Do not convert the numbers.
(a) 0x1F + 0xA9 1 1 F + A 9 C 8 ‾ (b) 0x3B – 0x12 3 B − 12 29 ‾
\begin{aligned}
\textrm{\textcolor{skyblue}{(a) 0x1F + 0xA9}}\qquad\qquad&\\[1em]
\overset{1}{1}F&\\
+A9&\\
\text{\textcolor{khaki}{ $\overline{C8}$}}& \\[2em]
\textrm{\textcolor{skyblue}{(b) 0x3B – 0x12}}\qquad\qquad&\\[1em]
3B&\\
-12&\\
\text{\textcolor{khaki}{ $\overline{29}$}}&\\[1em]
\end{aligned}
(a) 0x1F + 0xA9 1 1 F + A 9 C 8 (b) 0x3B – 0x12 3 B − 12 29
Fill in a truth table for the following functions. Show all steps.Mathematical Notation : x ∨ ( y ∧ z ) ⊕ y x \lor (y \wedge ~z) \oplus y x ∨ ( y ∧ z ) ⊕ y
x
y
z
¬ z \neg z ¬ z y ∧ ¬ z y \wedge \neg z y ∧ ¬ z x ∨ ( y ∧ ¬ z ) x \lor (y \wedge \neg z) x ∨ ( y ∧ ¬ z ) . . . ⊕ y ... \oplus y ... ⊕ y
1
1
1
0
0
1
0
1
1
0
1
1
1
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
0
1
1
0
0
0
1
0
1
0
1
1
1
0
0
0
1
0
0
0
0
0
0
0
1
0
1
1
(b) (x + y) & ~z xor yMathematical Notation : ( x ∨ y ) ∧ ¬ z ⊕ y (x \lor y) \wedge \neg z \oplus y ( x ∨ y ) ∧ ¬ z ⊕ y
x
y
z
¬ z \neg z ¬ z x ∨ y x \lor y x ∨ y ( x ∨ y ) ∧ ¬ z (x\lor y)\wedge \neg z ( x ∨ y ) ∧ ¬ z . . . ⊕ y ... \oplus y ... ⊕ y
1
1
1
0
1
0
1
1
1
0
1
1
1
0
1
0
1
0
1
0
0
1
0
0
1
1
1
1
0
1
1
0
1
0
0
0
1
0
1
1
1
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0